Detrimentally symmetrical cases:

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Detrimentally symmetrical cases: $0<G\leq g$

Two equal QDs create a symmetric system of levels in the ``horizontal'' sense: $\gamma_{E1}=\gamma_{E2}=\gamma$ and $P_{E1}=P_{E2}=P$ ( $\Gamma_1=\Gamma_2$ but $\gamma_\mathrm{TOT}\neq P_\mathrm{TOT}$ ). Effectively, the populations behave as if the two dots were uncoupled

$\displaystyle n_1=n_2=\frac{P}{\gamma+P}\,,\quad n_B= n_1 n_2\,,\quad n_{12}=0\,,$

(4.40)

although the system is always in strong coupling (FSC) with

$\displaystyle R=\frac{\vert\gamma-P\vert}{\gamma+P}g\,,$

(4.41)

and $D^s=2\sqrt{\gamma P}/(\gamma+P)$ . If $P\gg\gamma$ , or the other way around, the symmetry in the ``vertical'' sense is completely broken and

The second possibility is the case $\gamma_{E1}=P_{E1}$ and $\gamma_{E2}=P_{E2}$ ( $\Gamma_1\neq\Gamma_2$ but $\gamma_\mathrm{TOT} = P_\mathrm{TOT}$ ), which is symmetrical in the vertical sense. This is equivalent to both 2LS in a thermal bath of infinite temperature, so that pump and decay become equivalent. Now $D^s=2\sqrt{\gamma_{E1}\gamma_{E2}}/(\gamma_{E1}+\gamma_{E2})$ and the populations are also effectively uncoupled, , , $n_{12}=0$ . The Rabi reads

$\displaystyle R=\frac{\vert\Gamma_-\vert}{\Gamma_+}\sqrt{g^2-\Gamma_+^2}\,,$

(4.42)

and gives the conditions for SC/WC regions in terms of $\Gamma_+$ . If $\gamma_{E1}\gg\gamma_{E2}$ or the other way around, the symmetry in the ``horizontal'' sense is broken and

In the particular case where all the parameters are equal to $\gamma$ , the symmetry is perfect in all senses (). Again, all the levels are equally populated ( , ) and uncoupled ( $n_{12}=0$ ) but the Rabi vanishes. The SC/WC regimes become conventional with $z_{1,2}=\sqrt{g^2-\gamma^2}$ . Some asymmetry in the parameters is needed to reach the SSC and MC regimes.

Elena del Valle 2009-10-11